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Suppose I have a matrix $$ M = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix}. $$

This has a one-dimensional non-trivial kernel $$ \text{ker}(M) = \text{span} \left \{ \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}\right\}. $$

Now consider the right shift operator $S[(x_1, x_2, \dots)] = (0, x_1, x_2, \dots)$ which is kind of an infinite dimensional version of the matrix $M$. This operator is also supposed to have a one dimensional non-trivial kernel. But I can't seem to find an element of it! It seems like we need the vector $x = (x_1, x_2, \dots)$ to be trivial if $S$ is to map it to the zero vector. So where am I going wrong and what is an actual element of the kernel of $S$?

Similarly for the left shift operator, $T[(x_1, x_2, \dots)] = (x_2, x_3, \dots)$, this operator which is the adjoint of the right shift operator is said to have a kernel of dimension one. What is an element of the kernel in this case?

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The right shift operator is injective so it has a trivial kernel. Namely, if $$S[(x_1,x_2,\dots)] = (0,x_1,x_2,\dots) = (0,0,0,\dots)$$ then $x_1 = x_2 = \dots = 0$. The left shift operator has a one-dimensional kernel given by $$\operatorname{span} \{ [(1,0,0,\dots)] \}.$$

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    $\begingroup$ If I then took the 'double' left shift operator operator $T[(x_1,x_2, x_3, x_4, \dots)] = (x_3, x_4 ,\dots)$, would this operator be even worse than the single left shift operator $S$ as it has a two dimensional kernel and hence is Fredholm of index $2$? Is it the case that operators with Fredholm index $\alpha > 0$ are all equally bad in the sense of invertibility no matter what the value of $\alpha$? Or is there something that can be done an operator of index $1$ that can't be done with an operator of index $2$? $\endgroup$
    – ManUtdBloke
    Jan 25, 2017 at 8:58
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    $\begingroup$ I don't really understand your question. Yes, the double shift has a two dimensional kernel and is surjective so it would be a Fredholm operator of index $2$ but you can also have a Fredholm operator of index $2$ which has a three dimensional kernel and a one-dimensional co-kernel. $\endgroup$
    – levap
    Jan 25, 2017 at 9:25

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